A charge q1 = 5 ìC is at the origin. A second charge q2 = -3 ìC is on the x-axis, 0.8 m from the origin. The electric field at a point on the y-axis, 0.5 m from the origin is:
I don't understand the question please help . Thanks for reading
Sure, let's break down the given information and find the electric field at the specified point.
We have two charges: q1 = 5 μC at the origin and q2 = -3 μC on the x-axis, 0.8 m from the origin. We are looking for the electric field at a point on the y-axis, 0.5 m from the origin.
To find the electric field at a point, we need to consider the electric forces due to both charges at that point.
The formula to calculate the electric field (E) at a given point is:
E = k * (q / r^2)
where k is the electrostatic constant (k ≈ 9.0 x 10^9 Nm^2/C^2), q is the charge creating the electric field, and r is the distance from the charge to the point where the electric field is being calculated.
Now, let's calculate the electric field created by q1 and q2 separately and then combine them to find the total electric field at the specified point.
Electric Field Created by q1:
q1 = 5 μC (positive charge at the origin)
r1 = 0.5 m (distance from the origin to the specified point on the y-axis)
k ≈ 9.0 x 10^9 Nm^2/C^2
E1 = k * (q1 / r1^2) = (9.0 x 10^9 Nm^2/C^2) * (5 x 10^-6 C) / (0.5 m)^2
Electric Field Created by q2:
q2 = -3 μC (negative charge on the x-axis, 0.8 m from the origin)
r2 = 0.5 m (distance from q2 to the specified point on the y-axis)
k ≈ 9.0 x 10^9 Nm^2/C^2
E2 = k * (q2 / r2^2) = (9.0 x 10^9 Nm^2/C^2) * (-3 x 10^-6 C) / (0.5 m)^2
Now, we can add the two electric fields to get the total electric field at the specified point:
E_total = E1 + E2
Substituting the calculated values for E1 and E2, we can find the numerical value for the total electric field.
Note: The direction of the electric field depends on the signs of the charges. The electric field due to a positive charge points away from it, while the electric field due to a negative charge points towards it. So, we need to consider the signs of E1 and E2 to determine the direction of the total electric field, if required.
I hope this step-by-step breakdown helps clarify the process for finding the electric field at the specified point. Let me know if you need any further assistance!
Of course! I can help you understand the question and find the answer.
In this question, we have two charges - q1 and q2. The charge q1 is placed at the origin (0,0), and q2 is located on the x-axis at a distance of 0.8 m from the origin.
The problem asks us to find the electric field at a point on the y-axis, which is 0.5 m away from the origin.
To find the electric field at a point, we need to calculate the electric force that would act on a positive test charge placed at that point. Mathematically, the electric field E at a point is given by the equation:
E = k * (q / r^2)
Where:
- E is the electric field,
- k is the electrostatic constant (9 × 10^9 N m^2/C^2),
- q is the charge,
- r is the distance from the charge to the point where you want to calculate the electric field.
Now, let's determine the electric field at the point on the y-axis, 0.5 m away from the origin.
First, calculate the electric field contribution from q1 at this point:
E1 = k * (q1 / r1^2)
where:
- q1 is the charge at the origin (5 µC),
- r1 is the distance from q1 to the point on the y-axis (0.5 m).
Second, calculate the electric field contribution from q2 at this point:
E2 = k * (q2 / r2^2)
where:
- q2 is the charge on the x-axis (-3 µC),
- r2 is the distance from q2 to the point on the y-axis (0.5 m).
Finally, add the electric field contributions from both charges to find the total electric field at the point on the y-axis:
E_total = E1 + E2
Now you have the necessary information and equations to find the electric field at the given point. You can substitute the values for q1, q2, r1, r2, and k into the equations to get the final answer.
q₁ = 5 μC =5•10⁻⁶ C,
q₂ = - 3 μC = - 3•10⁻⁶ C,
r₁ =0.5 m,
r₂ = 0.8 m.
k =9•10⁹ N•m²/C²
E₁=k•q₁/ r₁², (directed ↑)
E₂ = k•q₂/ r₂², (directed ↑)
vector E= vector E₁+vectorE₂
The magnitude of vector E is directed ↗ and is calculated using the cosine law.
tanα =0.8/0.5=1.6
α =58⁰
E=sqrt{E₁²+E₂²-2E₁E₂cosα} = …